Mister Exam
(x^2-7|x|+10)/(x^2-6x+9)>=0 inequation
The solution
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[src]
2 x - 7*|x| + 10 --------------- >= 0 2 x - 6*x + 9
$$\frac{\left(x^{2} - 7 \left|{x}\right|\right) + 10}{\left(x^{2} - 6 x\right) + 9} \geq 0$$
(x^2 - 7*|x| + 10)/(x^2 - 6*x + 9) >= 0
Detail solution
Given the inequality:
$$\frac{\left(x^{2} - 7 \left|{x}\right|\right) + 10}{\left(x^{2} - 6 x\right) + 9} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(x^{2} - 7 \left|{x}\right|\right) + 10}{\left(x^{2} - 6 x\right) + 9} = 0$$
Solve:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = 5$$
$$x_{4} = -5$$
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = 5$$
$$x_{4} = -5$$
This roots
$$x_{4} = -5$$
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = 5$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{4}$$
For example, let's take the point
$$x_{0} = x_{4} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$-5.1$$
substitute to the expression
$$\frac{\left(x^{2} - 7 \left|{x}\right|\right) + 10}{\left(x^{2} - 6 x\right) + 9} \geq 0$$
$$\frac{\left(- 7 \left|{-5.1}\right| + \left(-5.1\right)^{2}\right) + 10}{9 + \left(\left(-5.1\right)^{2} - - 5.1 \cdot 6\right)} \geq 0$$
one of the solutions of our inequality is:
$$x \leq -5$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq -5$$
$$x \geq -2 \wedge x \leq 2$$
$$x \geq 5$$
$$\frac{\left(x^{2} - 7 \left|{x}\right|\right) + 10}{\left(x^{2} - 6 x\right) + 9} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(x^{2} - 7 \left|{x}\right|\right) + 10}{\left(x^{2} - 6 x\right) + 9} = 0$$
Solve:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = 5$$
$$x_{4} = -5$$
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = 5$$
$$x_{4} = -5$$
This roots
$$x_{4} = -5$$
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = 5$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{4}$$
For example, let's take the point
$$x_{0} = x_{4} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$-5.1$$
substitute to the expression
$$\frac{\left(x^{2} - 7 \left|{x}\right|\right) + 10}{\left(x^{2} - 6 x\right) + 9} \geq 0$$
$$\frac{\left(- 7 \left|{-5.1}\right| + \left(-5.1\right)^{2}\right) + 10}{9 + \left(\left(-5.1\right)^{2} - - 5.1 \cdot 6\right)} \geq 0$$
0.00472488949855208 >= 0
one of the solutions of our inequality is:
$$x \leq -5$$
_____ _____ _____
\ / \ /
-------•-------•-------•-------•-------
x4 x1 x2 x3Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq -5$$
$$x \geq -2 \wedge x \leq 2$$
$$x \geq 5$$
Solving inequality on a graph
Rapid solution
[src]
Or(And(-2 <= x, x <= 2), And(5 <= x, x < oo), And(x <= -5, -oo < x))
$$\left(-2 \leq x \wedge x \leq 2\right) \vee \left(5 \leq x \wedge x < \infty\right) \vee \left(x \leq -5 \wedge -\infty < x\right)$$
((-2 <= x)∧(x <= 2))∨((5 <= x)∧(x < oo))∨((x <= -5)∧(-oo < x))
Rapid solution 2
[src]
(-oo, -5] U [-2, 2] U [5, oo)
$$x\ in\ \left(-\infty, -5\right] \cup \left[-2, 2\right] \cup \left[5, \infty\right)$$
x in Union(Interval(-oo, -5), Interval(-2, 2), Interval(5, oo))